= Solution
Every counted point has <integer> coordinates and lies in the square $[-n,n]^2$, while the origin is always counted. Hence
$$
1\le a_n\le(2n+1)^2.
$$
No precise circle-area estimate is needed; this elementary bound controls the growth of the <power series> coefficients.
The <comparison test for series> states that if $0\le u_n\le v_n$, convergence of $\sum v_n$ implies convergence of $\sum u_n$. Equivalently, divergence of the smaller nonnegative series forces divergence of the larger one.
For $|z|=r<1$,
$$
\sum_{n=0}^\infty|a_nz^n|
\le\sum_{n=0}^\infty(2n+1)^2r^n.
$$
For $0<r<1$, the ratio of successive terms on the right tends to $r<1$, so the <ratio test> and the <comparison test for series> show <absolute convergence>. The case $r=0$ is immediate.
If $|z|\ge1$, then $|a_nz^n|\ge1$, so the terms do not tend to zero. The <term test for divergence> therefore rules out convergence, including every point on the boundary $|z|=1$. Thus the <radius of convergence> is
$$
\boxed{R=1.}
$$
\b[The series converges exactly in the open unit disc.] This is an example of <polynomial coefficient growth with unit power-series radius>: the lattice-count bound is polynomial rather than exponential.
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